# Practice Paper 2 Question 5

Find the gradient of the implicitly defined curve \(y^2+2y=x^3+7x\) at all its intersection points with the line \(x=1\).

## Related topics

## Hints

- Hint 1Can you differentiate both sides with respect to \(x.\)
- Hint 2How can you rearrange your equation to find \(\frac{dy}{dx}?\)
- Hint 3Try to substitute \(x=1\) into the original equation.

## Solution

By differentiating both sides with respect to \(x,\) we have \(2y\frac{dy}{dx}+2\frac{dy}{dx}=3x^2+7\) (using the chain rule). Rearranging gives \(\frac{dy}{dx}=\frac{3x^2+7}{2y+2}.\) Now we need to find all intersection points with the line \(x=1.\) These are given by \(y^2+2y-8=0,\) which has solutions \(y\in\{-4,2\}.\) We then plug this into the derivative expression to get the gradient values \(-\frac{5}{3}\) and \(\frac{5}{3}.\)

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at oi.[email protected]. Please do not write to this address regarding general admissions or course queries.